Question:
let sequence $\{x_{n}\}$ such $$x_{0}=0,x_{1}=1,x_{2}=0,x_{3}=1$$
and such $$x_{n+3}=\dfrac{(n^2+n+1)(n+1)}{n}x_{n+2}+(n^2+n+1)x_{n+1}-\dfrac{n+1}{n}x_{n}$$
show that:$ x_{n}$ are all square number?
My idea: I have $$x_{4}=3\cdot 2-2\cdot 1=4=2^2$$ $$x_{5}=\dfrac{21}{2}\cdot 4+(2^2+2+1)\cdot 1=49=7^2$$ and so on,but for all $n$,How prove it?
Thank you
In fact $x_{n+1}=a_n^2$, where $\{a_n\}$ are determined by the recurrence relation $$a_{n+1}=n a_n+a_{n-1}$$ with the initial conditions $a_0=1$, $a_1=0$. This can be exprerimentally discovered using the online encyclopedia of integer sequences, and then the proof can be easily carried out by induction.
Proof: The only nontrivial induction step: \begin{align} x_{n+3}=\frac{(n^2+n+1)(n+1)}{n}a_{n+1}^2+(n^2+n+1)a_{n}^2-\frac{n+1}{n}a_{n-1}^2=\ldots=a_{n+2}^2, \end{align} where the dots denote expressing $a_n$ and $a_{n-1}$ in terms of $a_{n+2}$ and $a_{n+1}$. $\quad \blacksquare$